Find the number of times digit 3 occurs in each and every number from 0 to n
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Find the number of times digit 3 occurs in each and every number from 0 to n
Program to find the number of times digit 3 occurs in each and every number from 0 to n is discussed here. Given a number n as input, count the number of 3s occurring in all the numbers from 0 to n.
For example,
Input: 100
Output: 20
Total number of 3s that appear from numbers 0 to 100 are {3, 13, 23, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 43, 53, 63, 73, 83, 93}
Algorithm to find the number of times digit 3 occurs in each and every number
Input the number n from the user.
Initialize count = 0.
Repeat for all numbers from o to n.
Find if num == 3
Increment count
Divide the number by 10 and from step 4.
Return count.
Program to find the number of times digit 3 occurs in each and every number from 0 to n is given below.
C
// C program to find the number of times digit 3 occurs in each and every number from 0 to n
#include
int count_3s(int n)
{
int count = 0;
while (n > 0)
{
if (n == 3)
{
count++;
}
n = n / 10;
}
return count;
}
int count_in_range(int n)
{
int count = 0 ;
for (int i = 2; i <= n; i++)
{
count += count_3s(i);
}
return count;
}
int main()
{
int n;
printf(“\nEnter the end value : “);
scanf(“%d”, &n);
printf(“\nTotal occurrences of 3 from 0 to %d is %d\n”, n,count_in_range(n));
return 0;
}
C++
// C++ program to find the number of times digit 3 occurs in each and every number from 0 to n
#include
using namespace std;
int count_3s(int n)
{
int count = 0;
while (n > 0)
{
if (n == 3)
{
count++;
}
n = n / 10;
}
return count;
}
int count_in_range(int n)
{
int count = 0 ;
for (int i = 2; i <= n; i++)
{
count += count_3s(i);
}
return count;
}
int main()
{
int n;
cout << “nEnter the end value : “;
cin >> n;
cout << “nTotal occurrences of 3 from 0 to ” << n>
return 0;
}
JAVA 8
// Java program to find the number of times digit 3 occurs in each and every number from 0 to n
import java.util.*;
public class Main
{
static int count_3s(int n)
{
int count = 0;
while (n > 0)
{
if (n == 3)
{
count++;
}
n = n / 10;
}
return count;
}
static int count_in_range(int n)
{
int count = 0 ;
for (int i = 2; i <= n; i++)
{
count += count_3s(i);
}
return count;
}
public static void main(String[] args)
{
int number;
Scanner sc = new Scanner(System.in);
System.out.print(“nEnter the number : “);
number = sc.nextInt();
System.out.print(count_in_range(number));
}
}
PYTHON 3
// Java program to find the number of times digit 3 occurs in each and every number from 0 to n
import java.util.*;
public class Main
{
static int count_3s(int n)
{
int count = 0;
while (n > 0)
{
if (n == 3)
{
count++;
}
n = n / 10;
}
return count;
}
static int count_in_range(int n)
{
int count = 0 ;
for (int i = 2; i <= n; i++)
{
count += count_3s(i);
}
return count;
}
public static void main(String[] args)
{
int number;
Scanner sc = new Scanner(System.in);
System.out.print(“nEnter the number : “);
number = sc.nextInt();
System.out.print(count_in_range(number));
}
}
OUTPUT
Enter the end value : 100
Total occurrences of 3 from 0 to 100 is 20
Time complexity: O(n)
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